# 2004 JAMB Mathematics Past Questions & Answers - page 1

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1
An arc of a circle of length 22 cm subtends an angle of 3xo at the center of the circle. Find the value of x if the diameter of the circle is 14 cm
A 60o
B 120o
C 180o
D 30o

Correct Option: A

Solution
$$ARC\hspace{1mm}length = (\frac{\theta}{360})\times 2\pi r\22=\frac{3x}{360}\times \left(2 \times(\frac{22}{7})\times(\frac{7}{1})\right)\3x = 180\x = \frac{180}{3}\x = 60^{\circ}$$

2
Find the value of α2 + β2 if α + β = 2 and the distance between points (1, α) and (β, 1)is 3 units
A 14
B 3
C 5
D 11

Correct Option: D

Solution
$$PQ = \sqrt{(β - 1)^{2} + (1 - α)^{2}}\\ 3 =\sqrt{(β^{2} -2β^{2} + 1 + 1 - 2α + α^{2})}\\ 3 = \sqrt{(α^{2} + β^{2} - 2α + 2β + 2)}\\ 3 = \sqrt{(α^{2} + β^{2} - 2(α + β) + 2)}\\ 3 = \sqrt{(α^{2} + β^{2} - 2 * 2 + 2)}\\ 3 = \sqrt{(α^{2} + β^{2} - 2)}\\ 9 = (α^{2} + β^{2} - 2)\\ α^{2} + β^{2} = 9 + 2\\ α^{2} + β^{2} = 11$$

3
The sum of the interior angles of a pentagon is 6x + 6y. Find y in the terms of x
A y = 90 - x
B y = 150 - x
C y = 60 - x
D y = 120 -x

Correct Option: A

Solution
6x + 6y = (n - 2) 180
6x + 6y= (5 - 2) 180
6(x + y) = 3 * 180
x + y = (3 * 180)/6
x + y = 90o
y = 90 - x

4
In the diagram above, PQ = 4 cm and TS = 6 cm. If the area of parallelogram PQTU is 32 cm 2, find the area of the trapezium PQRU
A 60 cm2
B 72 cm2
C 48 cm2
D 24 cm2

Correct Option: B

Solution
Area of parallelogram PQTU = base * height
32 = 4 * h
w = 32/4
w = 8
∴ = Area of Trapezium PQRU = (1/2)(4 + 14) *8
= 1/2 * (18 * 8)
= 72 cm2

5
Find the midpoint of the line joining P(-3, 5) and Q(5, -3).
A (1, 1)
B (2, 2)
C (4, 4)
D (4, -4)

Correct Option: A

Solution
$$Mid point = \frac{(x_1 + x_2)}{2} ; \frac{(y_1 + y_2)}{2}\\ = \frac{(-3 + 5)}{2} ; \frac{(5 - 3)}{2}\\ = \frac{2}{2} ; \frac{2}{2}\\ = (1, 1)$$

6
Determine the locus of a point inside a square PQRS which is eqidistant from PQ and QR
A The diagonal QS
B the perpendicular bisector of PQ
C The diagonal PR
D side SR

Correct Option: A

Solution
The diagonal QS bisects the angle formed by PQ and QR
∴ [A]

7
Find the value of x in the figure above
A 15√6
B 20√6
C 3√6
D 5√6

Correct Option: D

Solution
$$\frac{X}{sin45}=\frac{15}{sin60}\X=\frac{15sin45}{sin60}\X=\frac{15\times(\frac{1}{\sqrt{2}})}{\sqrt{\frac{3}{2}}}\X=\frac{15}{\sqrt{2}}\times\frac{2}{\sqrt{3}}\X=\frac{30}{\sqrt{6}}\X=\frac{30}{\sqrt{6}}\times\frac{\sqrt{6}}{\sqrt{6}}=\frac{30\sqrt{6}}{6}=5\sqrt{6}$$

8
P, R and S lie on a circle center as shown above while Q lies outside the circle. Find ∠PSO
A 45o
B 55o
C 35o
D 40o

Correct Option: C

Solution
xo = 35 + 29 (Exterior angle = sum of two interior opposite angles)
x = 55o (∠ at the center twice ∠ at circumference) y = 110o
∠PSO = ∠SPO (base ∠S of 1sc Δ b/c PO = SO)
∴ ∠PSO = (180 - 110)/2
= 35o

9
The locus of a point which is 5 cm from the line LM is a
A line distance 10 cm from LM and parallel to LM
B pair of line on opposite sides of LM and parallel to it, each distance 5 cm from LM
C line parallel to LM and 5 cm from LM
D pair of parallel lines on one side of LM and parallel to LM

Correct Option: B

Solution
The locus of a point which is 5cn from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM

10
If y = 3 cos(x/3), find dy/dx when x = (3π/2)
A 1
B -3
C 2
D -1

Correct Option: D

Solution
y = 3cos(x/3)
dy/dx = 3x(1/3)x - sin (x/3)
= - sin x/3
But x = 3π/2 ∴ -sin(x/3) = -sin(3π/6)
= -sin (3 * 180)/6
= - sin 90
= -1

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